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\title{Chapter 13: External Products}
\author{SCC ET AL}
%\institute[XX大学]{XX大学\quad 数学与统计学院\quad 数学与应用数学专业}
%\date{2025年6月}

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% 目录页
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% Section 0
%\section{INTRO.}
\begin{frame}{intro. }
    
In this chapter we discuss the simplest operation to be performed on {\color{red}modules over the Weyl algebra}: the external product. 

We also discuss the computation of the dimensions of the external product from the dimension of its factors. 

In particular, we will show that the external product of {\color{red}holonomic modules} is itself holonomic.

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
% Section 1
%\section{A.}
\begin{frame}[allowframebreaks]{A. }

\section{External Products of Algebras}

We begin with a general definition. 

Let $A, B$ be $K$-algebras. 

The tensor product $A \otimes_K B$ is a $K$-vector space, on which we define a {\color{red}multiplication}. 

For $a, a{\,}' \in A$ and $b, b{\,}' \in B$, let
\begin{equation}
(a \otimes b)(a{\,}' \otimes b{\,}') = aa{\,}' \otimes bb{\,}'.
\end{equation}

It is routine to check that $A \otimes_K B$ with this product is a $K$-algebra. 

This is called the {\color{red}external product} of $A$ and $B$. 

Since we use this construction very often, it is convenient to have a special notation for it: $A \widehat{\otimes} B$.

If we apply the construction to polynomial rings or to the Weyl algebra we do not get anything new: that is the secret of its power. 

The key result is the following theorem.

\textbf{Theorem 1.1.}
Let $R$ be a $K$-algebra and let $A$ and $B$ be subalgebras of $R$. 

Suppose that

\begin{enumerate}
    \item $R = AB$,
    \item $[A, B] = 0$,
    \item there exist $K$-bases $\{a_i : i \in \mathbb{N}\}$ and $\{b_j : j \in \mathbb{N}\}$ of $A$ and $B$, respectively, such that $\{a_i b_j : i, j \in \mathbb{N}\}$ is a $K$-basis of $R$.
\end{enumerate}

Then $R \cong A \widehat{\otimes} B$.


\textbf{Proof:} Define a map $\phi : A \times B \to R$ by $\phi(a, b) = ab$. 

It is clearly $K$-bilinear and $K$-balanced. 

Therefore, there exists a map $\Phi : A \otimes_K B \to R$ such that $\Phi(a \otimes b) = ab$. 

This is called the {\color{red}multiplication map}.


Let $a, a{\,}' \in A$ and $b, b{\,}' \in B$. Then

\begin{equation}
\Phi((a \otimes b)(a{\,}' \otimes b{\,}')) = \Phi(aa{\,}' \otimes bb{\,}') = aa{\,}'bb{\,}'.
\end{equation}

On the other hand,

\begin{equation}
\Phi(a \otimes b)\Phi(a{\,}' \otimes b{\,}') = aba{\,}'b{\,}'.
\end{equation}

Thus (2) implies that $\Phi$ is a $K$-algebra homomorphism.

Since, from (1), every element of $R$ is a linear combination of monomials $ab$, with $a \in A$ and $b \in B$, we conclude that $\Phi$ is surjective. 

The injectivity is a consequence of (3) and Corollary 12.4.3.

A carefully chosen notation makes the application of Theorem 1.1 to polynomial rings and the Weyl algebra almost tautological. 

Our choice is the following. 

Let $K[X] = K[x_1, \ldots, x_n]$ and $K[Y] = K[y_1, \ldots, y_m]$ be polynomial rings. 

Write $K[X,Y]$ for the polynomial ring on the $x$'s and $y$'s. 

Then $A_n$ will be the Weyl algebra generated by the $x$'s and $\partial_x$'s, and $A_m$ the Weyl algebra generated by the $y$'s and $\partial_y$'s. 

Both are subalgebras of $A_{m+n}$, the Weyl algebra generated by the $x$'s, $y$'s and their derivatives. 

We shall retain this notation for the rest of the chapter.

\newpage

\textbf{Corollary 1.2.}

The following isomorphisms are induced by the multiplication map:

\begin{enumerate}
    \item $K[X] \widehat{\otimes} K[Y] \cong K[X,Y]$.
    \item $A_m \widehat{\otimes} A_n \cong A_{m+n}$.
\end{enumerate}

The isomorphisms defined in the corollary are so natural that we shall take them to be equalities. 

Thus, from now on, we shall use without further comment that $A_m \widehat{\otimes} A_n = A_{m+n}$. 

In other words if $a \in A_n$ and $b \in A_m$, we will identify the monomial $ab$ with $a \otimes b$. 

Similarly for polynomial rings.


\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\section{2.}
\section{External Products of Modules}

\begin{frame}[allowframebreaks]{B. }

Once again we turn to the general situation. 

Let $A,B$ be $K$-algebras. 

Suppose that $M$ is a left $A$-module and that $N$ is a left $B$-module. 

Then we may turn the $K$-vector space $M \otimes_K N$ into a left $(A \widehat{\otimes} B)$-module. 

The action of $a \otimes b \in A \widehat{\otimes} B$ on $u \otimes v \in M \otimes_K N$ is given by the formula

\begin{equation}
(a \otimes b)(u \otimes v) = au \otimes bv.
\end{equation}

It is routine to check that $M \otimes_K N$ is a module for this action. 

We shall write $M \widehat{\otimes} N$ for this $(A \widehat{\otimes} B)$-module.

\newpage

\textbf{Proposition 2.1.}

If $M$ and $N$ are finitely generated modules, then so is $M \widehat{\otimes} N$.

Proof: Suppose that $M$ is generated by $u_1, \ldots, u_s$ over $A$ and that $N$ is generated over $B$ by $v_1, \ldots, v_t$. 

The elements of $M \widehat{\otimes} N$ are $K$-linear combinations of elements of the form $u \otimes v$, for $u \in M$, $v \in N$. 

But $u = \sum_{i=1}^{s} a_i u_i$ and $v = \sum_{j=1}^{t} b_j v_j$. Thus,

\begin{equation}
u \otimes v = \sum_{i,j} (a_i \otimes b_j)(u_i \otimes v_j).
\end{equation}

Hence $M \widehat{\otimes} N$ is generated over $A \widehat{\otimes} B$ by $u_i \otimes v_j$, for $1 \leq i \leq s$ and $1 \leq j \leq t$.

Now let $M$ be a left $A_m$-module and $N$ a left $A_n$-module. 

The external product $M \widehat{\otimes} N$ is a left $A_{m+n}$-module, under the convention that $A_{m+n} = A_m \widehat{\otimes} A_n$. 

This will be used so often, that even at the risk of being {\color{red}pedantic}, we write the action explicitly. Let $u \otimes v \in M \widehat{\otimes} N$. 

A monomial of $A_{m+n}$ can be written, in only one way, in the form $ab$ with $a \in A_m$ and $b \in A_n$. Thus,

\begin{equation}
(ab)(u \otimes v) = au \otimes bv.
\end{equation}

The situation for polynomial rings is entirely similar. 

The next lemma will be very useful in calculations. 

We make use of these conventions in its statement and proof.

\newpage

\textbf{Lemma 2.2.}

Let $\mathcal{I}$ be a left ideal of $A_m$ and $\mathcal{J}$ a left ideal of $A_n$. 

Denote by $A_{m+n} \mathcal{I} + A_{m+n} \mathcal{J}$ the left ideal of $A_{m+n}$ generated by the elements of $\mathcal{I}$ and $\mathcal{J}$. 

Then
\begin{equation}
(A_m/\mathcal{I}) \widehat{\otimes} (A_n/\mathcal{J}) \cong A_{m+n}/(A_{m+n} \mathcal{I} + A_{m+n} \mathcal{J})
\end{equation}

as $A_{m+n}$-modules.

Proof: Since every monomial in $A_{m+n}$ may be written, uniquely, in the form $ab$, for $a \in A_m$, $b \in A_n$, there exists a $K$-linear map,

\begin{equation}
\psi : A_{m+n} \to (A_m/\mathcal{I}) \widehat{\otimes} (A_n/\mathcal{J})
\end{equation}


given by $\psi(ab) = (a + \mathcal{I}) \otimes (b + \mathcal{J})$. 

An easy calculation shows that this is an $A_{m+n}$-module homomorphism. 

It is clearly surjective. 

On the other hand, a monomial $ab$ with $a \in \mathcal{I}$ or $b \in \mathcal{J}$, gives $\psi(ab) = 0$. 

Hence $A_{m+n}\mathcal{I} + A_{m+n}\mathcal{J} \subseteq \ker \psi$.

By the {\color{red}universal property of the tensor product}, there is a $K$-linear map,

\begin{equation}
\phi : (A_m/\mathcal{I}) \widehat{\otimes} (A_n/\mathcal{J}) \to A_{m+n}/(A_{m+n}\mathcal{I} + A_{m+n}\mathcal{J}),
\end{equation}

defined by $\phi((a+\mathcal{I}) \otimes (b+\mathcal{J})) = ab + A_{m+n}\mathcal{I} + A_{m+n}\mathcal{J}$. 

This is a surjective map, and

\begin{equation}
\phi\psi(ab) = ab + A_{m+n}\mathcal{I} + A_{m+n}\mathcal{J}.
\end{equation}

Hence $A_{m+n}\mathcal{I} + A_{m+n}\mathcal{J} = \ker \psi$ and $\psi$ is an isomorphism of $A_{m+n}$-modules.

Although we have not used in the proof of 2.2 that $\phi$ is a homomorphism of $A_{m+n}$-modules, this is easily shown to be true.

\newpage

\textbf{Corollary 2.3.}

Let $\mathcal{I}$ be a left $A_m$-module. 

Then
\begin{equation}
(A_m/\mathcal{I}) \widehat{\otimes} A_n \cong A_{m+n}/A_{m+n}\mathcal{I}
\end{equation}
is an isomorphism of $A_{m+n}$-$A_n$-bimodules.

Proof: It follows from Lemma 2.2 that there exists an isomorphism of left $A_{m+n}$-modules,

\begin{equation}
\psi : A_{m+n}/A_{m+n}\mathcal{I} \to (A_m/\mathcal{I}) \widehat{\otimes} A_n,
\end{equation}

given by $\psi(ab + A_{m+n}\mathcal{I}) = (a + \mathcal{I}) \otimes b$, where $a \in A_m$, $b \in A_n$. 

Note that since $\mathcal{I} \subseteq A_m$, the elements of $\mathcal{I}$ commute with the elements of $A_n$. 

Hence $A_{m+n}/A_{m+n}\mathcal{I}$ is a right $A_n$-module. 

So is $(A_m/\mathcal{I}) \widehat{\otimes} A_n$, with $A_n$ acting on the right component of the tensor product. 

Let $c \in A_n$. 

Then
\begin{equation}
\psi((ab)c + A_{m+n}\mathcal{I}) = (a + \mathcal{I}) \otimes bc = \psi(ab + A_{m+n}\mathcal{I})c.
\end{equation}

Therefore, $\psi$ is also a right $A_n$-module isomorphism.


\end{frame}

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%\section{3.}
\section{Graduations and Filtrations}
\begin{frame}[allowframebreaks]{C. }

We will now construct a {\color{red}good filtration} for the external product of modules over the Weyl algebra and calculate its associated graded module. 

Throughout this section, let $M$ be a finitely generated left $A_m$-module with good filtration $\Gamma(M) = \{\Gamma_i(M): i \in \mathbb{N}\}$ and let $N$ be a finitely generated left $A_n$-module with good filtration $\Gamma(N)$. 

We will proceed through a series of lemmas.

\newpage

\textbf{Lemma 3.1.}

The Bernstein filtration of $A_{m+n}$ satisfies

\begin{equation}
B_t(A_{m+n}) = \sum_{p+q=t} B_p(A_m)B_q(A_n).
\end{equation}

Proof: First note that

\begin{equation}
B_p(A_m)B_q(A_n) \subseteq B_{p+q}(A_{m+n}).
\end{equation}

Now every monomial in $B_t(A_{m+n})$ may be written as a product $ab$, with $a \in A_m$ and $b \in A_n$. Assume that $a,b$ have degrees $p,q$, respectively. 

Since $ab$ has degree $p+q$, it follows that $ab \in B_p(A_m)B_q(A_n)$. 

But an element of $B_t(A_{m+n})$ is a sum of such monomials. 

Therefore, $B_t(A_{m+n}) = \sum_{p+q=t} B_p(A_m)B_q(A_n)$.

\newpage

\textbf{Lemma 3.2.}

The $K$-vector spaces
\begin{equation}
\Gamma_k(M \widehat{\otimes} N) = \sum_{i+j=k} \Gamma_i(M) \otimes \Gamma_j(N)
\end{equation}
form a good filtration of $M \widehat{\otimes} N$ as an $A_{m+n}$-module.

Note that we are identifying $\Gamma_i(M) \otimes_K \Gamma_j(N)$ with a subspace of $M \widehat{\otimes} N$, in the obvious way.

Proof: Since the summation in the definition of $\Gamma_k(M \widehat{\otimes} N)$ is finite, we have by Corollary 12.4.3 that $\Gamma_k(M \widehat{\otimes} N)$ is a finite dimensional $K$-vector space.

Now every element of $M \widehat{\otimes} N$ is a finite linear combination of elements of the form $u \otimes v$, with $u \in M$, $v \in N$. 

Assume that $u \in \Gamma_i(M)$ and $v \in \Gamma_j(N)$. Then
\begin{equation}
u \otimes v \in \Gamma_i(M) \otimes \Gamma_j(N) \subseteq \Gamma_{i+j}(M \widehat{\otimes} N).
\end{equation}

Hence, $M \widehat{\otimes} N = \bigcup_{k \geq 0} \Gamma_k(M \widehat{\otimes} N)$.

Finally, since
\begin{equation}
(B_p(A_m)B_q(A_n))(\Gamma_i(M) \otimes_K \Gamma_j(N)) \subseteq \Gamma_{i+p}(M) \otimes_K \Gamma_{j+q}(N),
\end{equation}
we conclude, using Lemma 3.1, that 
$$
B_t(A_{m+n})\Gamma_k(M \widehat{\otimes} N) = \Gamma_{k+t}(M \widehat{\otimes} N) \,\text{for}\, t \gg 0. 
$$ 

Thus $\Gamma_k(M \widehat{\otimes} N)$ is a good filtration of $M \widehat{\otimes} N$.

We will now calculate the graded module with respect to this good filtration. 

The ring $S_n = gr^F A_n$ is a polynomial ring by Theorem 7.3.1. 

Using the conventions of §1 we may identify $S_m \widehat{\otimes} S_n$ with $S_{m+n}$. 

Denote the homogeneous component of degree $t$ of $S_n$ by $S_n(t)$. 

By an argument similar to that used in Lemma 3.1, we have that $$S_{n+m}(t) = \bigoplus_{p+q=t} S_p(A_m)S_q(A_n). $$

Finally, write $gr_k(M) = \Gamma_k(M)/\Gamma_{k-1}(M)$.

\newpage

\textbf{Lemma 3.3.}

There is an isomorphism of $K$-vector spaces
\begin{equation}
gr_k(M \widehat{\otimes} N) \cong \bigoplus_{i+j=k} gr_i(M) \otimes_K gr_j(N).
\end{equation}

\textbf{Proof:} Consider first the $K$-linear map,
\begin{equation}
\eta_{ij}: \Gamma_i(M) \otimes_K \Gamma_j(N) \to gr_i(M) \otimes_K gr_j(N)
\end{equation}
defined by $\eta_{ij}(u \otimes v) = \mu_i(u) \otimes \mu_j(v)$, where we are denoting the {\color{red}symbol maps} of both $M$ and $N$ by $\mu$. 

It is easy to see that $\eta_{ij}$ is surjective and that its kernel is the subspace
\begin{equation}
\Gamma_{i-1}(M) \otimes \Gamma_j(N) + \Gamma_i(M) \otimes \Gamma_{j-1}(N).
\end{equation}

Putting these maps together, we get a linear map
\begin{equation}
\eta: \bigoplus_{i+j=k} (\Gamma_i(M) \otimes \Gamma_j(N)) \to \bigoplus_{i+j=k} (gr_i(M) \otimes gr_j(N)).
\end{equation}

Note that the canonical projection gives a map
\begin{equation}
\bigoplus_{i+j=k} (\Gamma_i(M) \otimes \Gamma_j(N)) \to \Gamma_k(M \widehat{\otimes} N).
\end{equation}

But $\ker \eta_{ij} \subseteq \Gamma_{i+j-1}(M \widehat{\otimes} N)$, thus
\begin{equation}
\bigoplus_{i+j=k} (\Gamma_i(M) \otimes \Gamma_j(N))/\bigoplus_{i+j=k} \ker \eta_{ij} \to \bigoplus_{i+j=k} (gr_i(M) \otimes gr_j(N)).
\end{equation}

factors through
\begin{equation}
\Gamma_k(M \widehat{\otimes} N)/\Gamma_{k-1}(M \widehat{\otimes} N) \to \bigoplus_{i+j=k} (gr_i(M) \otimes gr_j(N)).
\end{equation}

Since the former map is bijective, so is the latter, which gives the required isomorphism.

Taking direct sums for $k \geq 0$, and using Lemma 3.3, we conclude that there exists an isomorphism of $K$-vector spaces,
\begin{equation}
\theta : gr(M \widehat{\otimes} N) \to gr(M) \widehat{\otimes} gr(N).
\end{equation}

Since $M \widehat{\otimes} N$ is an $A_{m+n}$-module, $gr(M \widehat{\otimes} N)$ is an $S_{m+n}$-module. 

On the other hand, since $S_{m+n} = S_m \widehat{\otimes} S_n$, it follows that $gr(M) \widehat{\otimes} gr(N)$ is an $S_{m+n}$-module.

\newpage

\textbf{Theorem 3.4.}

The linear map $\theta$ is an isomorphism of $S_{m+n}$-modules.

Proof: We have only to check that $\theta$ is compatible with the action of $S_{m+n}$. 

Choose a monomial of $S_{m+n}$, and write it in the form $fg$, with $f \in S_m(p)$, $g \in S_n(q)$. There exist operators $a,b \in A_{m+n}$, such that

\begin{equation}
f = \sigma_p(a) \quad \text{and} \quad g = \sigma_q(b).
\end{equation}

In particular, $a \in A_m$ and $b \in A_n$. Now let $u \otimes v \in \Gamma_i(M) \otimes \Gamma_j(N)$, with $i + j = k$. Then

\begin{equation}
(fg)\mu_k(u \otimes v) = \mu_{k+p+q}((ab)(u \otimes v)).
\end{equation}

But $(ab)(u \otimes v) = au \otimes bv$; therefore

\begin{equation}
(fg)\mu_k(u \otimes v) = \mu_{k+p+q}(au \otimes bv).
\end{equation}

The latter is mapped onto $\mu_{i+p}(au) \otimes \mu_{j+q}(bv)$ by $\theta$. Since this may be rewritten as $(fg)(\mu_i(u) \otimes \mu_j(v))$, we have

\begin{equation}
\theta((fg)\mu_k(u \otimes v)) = (fg)(\mu_i(u) \otimes \mu_j(v))
\end{equation}

which completes the proof.


\end{frame}

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%\section{4.}
\section{Dimensions and Multiplicities}
\begin{frame}[allowframebreaks]{D. }


We are now ready to calculate the dimension and multiplicity of an external product.

\textbf{Theorem 4.1.}

Let $M$ be a finitely generated left $A_m$-module and $N$ a finitely generated left $A_n$-module. Then

(1) $d(M \widehat{\otimes} N) = d(M) + d(N)$,

(2) $m(M \widehat{\otimes} N) \leq m(M)m(N)$.

Proof: We retain the notation of §3. It follows from Lemma 3.3 that

\begin{equation}
\dim_K \Gamma_k(M \widehat{\otimes} N) = \sum_{r=0}^{k} \sum_{i+j=r} \dim_K gr_i(M) \dim_K gr_j(N).
\end{equation}

Therefore,

\begin{equation}
\dim_K \Gamma_k(M \widehat{\otimes} N) \leq \sum_{i=0}^{k} \dim_K gr_i(M) \sum_{j=0}^{k} \dim_K gr_j(N).
\end{equation}

From this inequality and Lemma 3.2 we have that

\begin{equation}
\dim_K \Gamma_k(M \widehat{\otimes} N) \leq \dim_K \Gamma_k(M) \dim_K \Gamma_k(N) \leq \dim_K \Gamma_{2k}(M \widehat{\otimes} N).
\end{equation}

Assuming that $k \gg 0$, and using the Hilbert polynomials of the corresponding filtrations we get that

\begin{equation}
\chi(k, M \widehat{\otimes} N) \leq \chi(k, M) \chi(k, N) \leq \chi(2k, M \widehat{\otimes} N).
\end{equation}

Since this holds for all large values of $k$, both (1) and (2) immediately follow.

The corollary is an easy consequence of the theorem.

\textbf{Corollary 4.2.}

Let $M$ be a holonomic $A_m$-module and $N$ a holonomic $A_n$-module. Then $M \widehat{\otimes} N$ is a holonomic $A_{m+n}$-module.


\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\section{5.}
\section{Exercises}

\begin{frame}[allowframebreaks]{E. }

\textbf{Exercise 5.1.}

Let $B_n$ be the ring of differential operators of the rational function field $K(X) = K(x_1, \ldots, x_n)$. Is it true that $B_n \widehat{\otimes} B_m \cong B_{n+m}$ as $K$-algebras?

\newpage

\textbf{Exercise 5.2.}

Show that the multiplication map induces an isomorphism of $A_{m+n}$-modules,

\begin{equation}
K[X,Y] \cong K[X] \widehat{\otimes} K[Y].
\end{equation}

\newpage

\textbf{Exercise 5.3.}

Let $p \in K[X]$ and $q \in K[Y]$. Show that there exists an isomorphism of $A_{m+n}$-modules,

\begin{equation}
K[X, Y(pq)^{-1}] \cong K[X, p^{-1}] \widehat{\otimes} K[Y, q^{-1}].
\end{equation}

\newpage

\textbf{Exercise 5.4.}

Let $\sigma$, $\sigma'$ be automorphisms of $A_m$ and $A_n$, respectively. Let $M$ be a left $A_m$-module and $N$ be a left $A_n$-module.

(1) Show that there exists a unique automorphism $\theta$ of $A_{m+n}$ whose restriction to $A_m$ is $\sigma$ and whose restriction to $A_n$ is $\sigma'$.

(2) Show that
$%\begin{equation}
M_\sigma \widehat{\otimes} N_{\sigma'} \cong (M \widehat{\otimes} N)_\theta.
$%\end{equation}

\newpage

\textbf{Exercise 5.5.}

Let $M$ be a left $A_m$-module and $N$ a left $A_n$-module. 

Show that the Fourier Transform satisfies
\begin{equation}
(M \widehat{\otimes} N)_\mathcal{F} \cong M_\mathcal{F} \widehat{\otimes} N_\mathcal{F}.
\end{equation}

\newpage

\textbf{Exercise 5.6.}

Let $M$ be a holonomic $A_m$-module and $N$ a holonomic $A_n$-module. 

Show that if $M$ and $N$ have multiplicity 1, then $M \widehat{\otimes} N$ is a {\color{red}simple holonomic} $A_{m+n}$-module.

\newpage

\textbf{Exercise 5.7.}

In this exercise we use the results of Ch. 11. 

Let $M$ be a left $A_m$-module and $N$ a left $A_n$-module. 

Show that $\text{Ch}(M \widehat{\otimes} N) = \text{Ch}(M) \times \text{Ch}(N)$.

\newpage

\textbf{Exercise 5.8.}

Let $1 \leq i \leq n$. Suppose that $d_i$ is an operator of $A_n$ of degree $\geq 1$ which is a linear combination of monomials in $x_i$ and $\partial_i$. 

Let $\mathcal{J}$ be the left ideal generated by $d_1, \ldots, d_n$. 

Show that $A_n/\mathcal{J}$ is a {\color{red}holonomic} module over $A_n$. 

Hint: $A_n/\mathcal{J} \cong (A_1/A_1d_1) \widehat{\otimes} \cdots \widehat{\otimes} (A_1/A_1d_n)$.

\end{frame}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%\section{6.}
\section{Study Notes}

\begin{frame}[fragile,allowframebreaks]{F. }

\textbf{Notes 6.1.}

Elaborate the diagram

\[
\begin{tikzcd}
A_1\text{-modules} \arrow[r, "\mathrm{Sol}"] \arrow[d, "\mathcal{F}"'] 
  & \text{Distributions} \arrow[d, "\mathcal{F}"] \\
A_1\text{-modules} \arrow[r, "\mathrm{Sol}"'] 
  & \text{Distributions}
\end{tikzcd}
\]

\end{frame}

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\end{document}


